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Expectation value proof

  1. Sep 24, 2007 #1
    I am trying to show that


    With the wavefunction [tex]\Psi[/tex] being both normalized to unity and square integrable

    Here is what I tried...

    [tex]<xp> = \int_{-\infty}^{\infty}{\Psi}^*xp{\Psi}dx[/tex]

    [tex]<px> = \int_{-\infty}^{\infty}{\Psi}^*px{\Psi}dx=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*\frac{\partial}{{\partial}x}(x{\Psi})dx=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*({\Psi}+x\frac{{\partial}{\Psi}}{{\partial}x})dx=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*{\Psi}dx+\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*x\frac{{\partial}{\Psi}}{{\partial}x}dx[/tex]

    from here the first part in the addition reduces to [tex]\frac{\hbar}{i}[/tex] since the integral reduces to 1 for a normalized wavefunction. The second part is just the integral for <xp>, so I'll just reduce it to that for now...

    [tex]<px>=\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*{\Psi}dx+\frac{\hbar}{i}\int_{-\infty}^{\infty}{\Psi}^*x\frac{{\partial}{\Psi}}{{\partial}x}dx =\frac{\hbar}{i}+<xp>[/tex]

    so now we have

    [tex]\frac{1}{m}(<xp>+<px>) = \frac{\hbar}{im}+\frac{2}{m}<xp>[/tex]............(2)

    Looking at the left side of of equation (1) now


    now.. I don't know if I can even do this: [tex]\frac{d}{dt}x^2=2x\frac{dx}{dt}=\frac{2}{m}xp[/tex]

    assuming I could, that reduces the second integral to [tex]\int_{-\infty}^{\infty}\Psi^*\frac{d}{dt}(x^2){\Psi}dx=\frac{2}{m}\int_{-\infty}^{\infty}\Psi^*xp{\Psi}dx=\frac{2}{m}<xp>[/tex]

    subbing into the LS and RS of equation (1) gives

    [tex]\int_{-\infty}^{\infty}x^2\frac{{\partial}}{{\partial}t}(\Psi^*{\Psi})dx + \frac{2}{m}<xp> = \frac{\hbar}{im} + \frac{2}{m}<xp>[/tex]

    the same terms on each side cancel leaving me to show

    [tex]\int_{-\infty}^{\infty}x^2\frac{{\partial}}{{\partial}t}(\Psi^*{\Psi})dx = \frac{\hbar}{im}[/tex]

    The inner derivative of this equation can be expanded to


    From Shroedinger's eq'n we know



    we then find that


    From here I don't know where I'm going..seems like kind of a dead end..
  2. jcsd
  3. Sep 25, 2007 #2


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    I'll give you a hint. Work in the Heisenberg representation. No integrations, just operator equalities (surely, assuming a common dense domain for the operators and a specific form of the Hamiltonian).
  4. Sep 25, 2007 #3
    I don't really know anything about 'Heisenberg representation'...this is homework too, so it would probably be pointless to figure out since we aren't expected to know it.
    Last edited: Sep 25, 2007
  5. Sep 26, 2007 #4


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    No, you can't; x doesn't depend on t, it's just an integration variable.

    Start with
    [tex]{\partial\over\partial t}\langle x^2\rangle=\int_{-\infty}^{\infty}\left[{\partial\Psi^*\over\partial t}x^2\Psi+\Psi^*x^2{\partial\Psi\over\partial t}\right]dx[/tex]
    Then use the Schrodinger equation and it's complex conjugate. Then do some integrations by parts to try to get it to look like your integral expression for [tex]\langle(xp+px)\rangle[/tex]

    Dexter is right that this problem is *much* easier in the Heisenberg representation.
  6. Sep 26, 2007 #5
    Thanks, I got it to work out now.
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